Problem: Jamal gets ready for a basketball game by shooting $10$ free-throws. Based on previous data, he has a $70\%$ chance of making each free-throw. Assume that the results of each free-throw are independent. Which of the following would find the probability of Jamal making exactly $8$ of $10$ free-throws? Choose 1 answer: Choose 1 answer: (Choice A) A ${10 \choose 7}(0.70)^7(0.30)^3$ (Choice B) B ${10 \choose 8}(0.70)^8(0.30)^2$ (Choice C) C ${10 \choose 8}(0.70)^2(0.30)^8$ (Choice D) D ${70 \choose 10}(0.70)^8(0.30)^2$ (Choice E) E $(0.70)^8(0.30)^2$
Answer: Probability of $8$ successes We want the probability that Jamal has $8$ successes (made free-throws) in $10$ trials (attempts), so we're going to need $2$ failures (missed free-throws) as well. The probability of each success is ${70\%}$ and the probability of each failure is $30\%}$. Since we were told to assume independence, we can multiply probabilities to find the probability of getting $8$ successes followed by $2$ failures: $\begin{aligned} P(\text{SSSSSSSSFF})&=({0.70})({0.70})\dots({0.70})(0.30})(0.30}) \\\\ &=({0.70})^8(0.30})^2 \end{aligned}$ The binomial coefficient ${n \choose k}$ SSSSSSSSFF isn't the only arrangement that produces $8$ successes in $10$ trials. For instance, FSSSSSSSSF would also produce the desired outcome. To count how many possible arrangements there are, we use the binomial coefficient ${n \choose k}$. It tells us the number of possible arrangements for $k$ successes in $n$ trials. In this problem, we want $k=8$ successes (made free-throws) in $n=10$ trials (attempts), so we should use the binomial coefficient ${10 \choose 8}$. [Tell me more about the binomial coefficient.] Putting it together Each arrangement has probability $(0.70)^8(0.30)^2$ so for our final answer we multiply this probability by the number of possible arrangements: ${10 \choose 8}(0.70)^8(0.30)^2$ The answer: ${10 \choose 8}(0.70)^8(0.30)^2$